How Fast Must a Person Run to Catch a Ball Dropped from a Building?

[wallace13]

A ball is thrown upward from the top of a 24.2-m-tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 29.2 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?



displacement (y)= Vi + at
displacement (x)/ time= Velocity

24.3 m = 12 + 9.81t
t= 1.24 s

29.2 m/ 1.24s = 23.54 m/s

By solving for time in the y-direction and then dividing the distance the person has to run by that time, I thought that i should get the correct velocity, but its wrong

 

[turin]

By solving for time in the y-direction and then dividing the distance the person has to run by that time, I thought that i should get the correct velocity,

Yes, however

displacement (y)= Vi + at

is incorrect. This equation is for v_f after some constant acceleration, not displacement.

Sketching a graph of y vs. t should also help.

 

[timmay]

The building is 24.2m tall, and the ball is thrown upwards. So it will reach some height above the building, before falling that height plus the height of the building before reaching the runner.

Try separating the y motion into two - firstly, the motion from release to the apex of the throw, and then the motion from apex of the throw to the ground.

displacement (y)= Vi + at

This equation is also wrong. Try looking here to see where you've gone wrong.